Let `alpha, beta` be any two positive values of x for which `2"cos"x, |"cos"x| " and " 1-3"cos"^(2) x` are in G.P. The minimum value of `|alpha -beta|`, is

Options

A. `pi//3`
B. `pi//4`
C. `pi//2`
D. none of these

Answer

To solve the problem, we need to find the minimum value of \(|\alpha - \beta|\) where \(\alpha\) and \(\beta\) are two positive values of \(x\) for which the terms \(2 \cos x\), \(|\cos x|\), and \(1 - 3 \cos^2 x\) are in geometric progression (G.P.).

### Step-by-Step Solution:

1. **Understanding the condition for G.P.**:
For three terms \(a\), \(b\), and \(c\) to be in G.P., the condition is:
\[ b^2 = ac \]
Here, let \(a = 2 \cos x\), \(b = |\cos x|\), and \(c = 1 - 3 \cos^2 x\).

2. **Setting up the equation**:
Applying the G.P. condition:
\[ |\cos x|^2 = (2 \cos x)(1 - 3 \cos^2 x) \]
Since \(\cos x\) is positive in the range we are considering, we can drop the absolute value:
\[ \cos^2 x = 2 \cos x (1 - 3 \cos^2 x) \]

3. **Rearranging the equation**:
Expanding the right-hand side:
\[ \cos^2 x = 2 \cos x - 6 \cos^3 x \]
Rearranging gives:
\[ 6 \cos^3 x + \cos^2 x - 2 \cos x = 0 \]

4. **Factoring out \(\cos x\)**:
Factoring out \(\cos x\):
\[ \cos x (6 \cos^2 x + \cos x - 2) = 0 \]
This gives us one solution:
\[ \cos x = 0 \quad \text{(not positive, discard)} \]
Now we solve the quadratic:
\[ 6 \cos^2 x + \cos x - 2 = 0 \]

5. **Using the quadratic formula**:
Applying the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here \(a = 6\), \(b = 1\), and \(c = -2\):
\[ \cos x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 6 \cdot (-2)}}{2 \cdot 6} \]
\[ = \frac{-1 \pm \sqrt{1 + 48}}{12} = \frac{-1 \pm 7}{12} \]
This gives:
\[ \cos x = \frac{6}{12} = \frac{1}{2} \quad \text{or} \quad \cos x = \frac{-8}{12} = -\frac{2}{3} \quad \text{(discard negative)} \]

6. **Finding the angles**:
The positive solution \(\cos x = \frac{1}{2}\) corresponds to:
\[ x = \frac{\pi}{3} \]
The other solution \(\cos x = -\frac{2}{3}\) gives:
\[ x = \cos^{-1}(-\frac{2}{3}) \]

7. **Calculating \(|\alpha - \beta|\)**:
Let \(\alpha = \frac{\pi}{3}\) and \(\beta = \cos^{-1}(-\frac{2}{3})\). We need to find:
\[ |\alpha - \beta| = \left| \frac{\pi}{3} - \cos^{-1}(-\frac{2}{3}) \right| \]

Since \(\cos^{-1}(-\frac{2}{3})\) is in the second quadrant, we can compute:
\[ \cos^{-1}(-\frac{2}{3}) = \pi - \cos^{-1}(\frac{2}{3}) \]

Therefore:
\[ |\alpha - \beta| = \left| \frac{\pi}{3} - \left(\pi - \cos^{-1}(\frac{2}{3})\right) \right| \]
\[ = \left| \frac{\pi}{3} - \pi + \cos^{-1}(\frac{2}{3}) \right| \]
\[ = \left| -\frac{2\pi}{3} + \cos^{-1}(\frac{2}{3}) \right| \]

8. **Finding the minimum value**:
The minimum value occurs when \(\alpha = \frac{\pi}{3}\) and \(\beta = \frac{2\pi}{3}\):
\[ |\alpha - \beta| = \left| \frac{\pi}{3} - \frac{2\pi}{3} \right| = \frac{\pi}{3} \]

### Final Answer:
The minimum value of \(|\alpha - \beta|\) is \(\frac{\pi}{6}\).